"Quantum modular forms"의 두 판 사이의 차이
둘러보기로 가기
검색하러 가기
imported>Pythagoras0 |
imported>Pythagoras0 |
||
| 18번째 줄: | 18번째 줄: | ||
U(-1;\zeta)=\sum_{n=0}^{k-1} (1+\zeta)^2(1+\zeta^2)^2\cdots (1+\zeta^n)^2\zeta^{n+1} | U(-1;\zeta)=\sum_{n=0}^{k-1} (1+\zeta)^2(1+\zeta^2)^2\cdots (1+\zeta^n)^2\zeta^{n+1} | ||
$$ | $$ | ||
| − | |||
* $R(-1;q)=f(q)$ and $C(-1;q)=b(q)$ in [[3rd order mock theta functions]] | * $R(-1;q)=f(q)$ and $C(-1;q)=b(q)$ in [[3rd order mock theta functions]] | ||
| + | * Thus if $\zeta$ be even $2k$ order root of unity | ||
| + | $$ | ||
| + | \lim_{q\to \zeta} f(q)-(-1)^k b(q)=-4\sum_{n=0}^{k-1} (1+\zeta)^2(1+\zeta^2)^2\cdots (1+\zeta^n)^2\zeta^{n+1} | ||
| + | $$ | ||
2013년 3월 18일 (월) 14:51 판
example
- unimodular generating function
$$ U(w;q)=\sum_{n=0}^{\infty}(wq;q)_{n}(w^{-1}q;q)_{n}q^{n+1} $$
$$R(w;q)=\sum_{n=0}^\infty \frac{q^{n^2}}{(wq;q)_n(w^{-1}q;q)_n}$$
$$C(w;q)=\frac{(q)_{\infty}}{(wq;q)_{\infty}(w^{-1}q;q)_{\infty}}$$
- limit formula $\zeta_b=e^{2\pi i/b}$, $1\le a <b$, for every root of unity $\zeta$, there exists an integer $c$ such that
$$ \lim_{q\to \zeta} R(\zeta_{b}^{a};q)-\zeta_{b^2}^{c} C(\zeta_{b}^{a};q)=-(1-\zeta_{b}^{a})(1-\zeta_{b}^{-a})U(\zeta_{b}^{a};\zeta) $$
special case
- If $b=2$ and $a=1$, then $\zeta_{b}^{a}=-1$
- $U(-1;\zeta)$ becomes a finite sum if $\zeta$ is a root of unity
$$ U(-1;\zeta)=\sum_{n=0}^{k-1} (1+\zeta)^2(1+\zeta^2)^2\cdots (1+\zeta^n)^2\zeta^{n+1} $$
- $R(-1;q)=f(q)$ and $C(-1;q)=b(q)$ in 3rd order mock theta functions
- Thus if $\zeta$ be even $2k$ order root of unity
$$ \lim_{q\to \zeta} f(q)-(-1)^k b(q)=-4\sum_{n=0}^{k-1} (1+\zeta)^2(1+\zeta^2)^2\cdots (1+\zeta^n)^2\zeta^{n+1} $$